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# Really confused, pls help asap

factorise each of the following expressions completely

a) 6x^2y^2-5xy-4

b) 3z-8xyz+4x^2y^2z

c) 36y^2-49(x+1)^2

### 1 Answer by Expert Tutors

Marina K. | The Beautiful MathThe Beautiful Math
4.7 4.7 (25 lesson ratings) (25)
2
If you rewrite the problems and substitute x*y by t for example, you'll get a quadratic equation.
for b)  part you have the same after factoring out z.
c) part is a2-bproblem.

a) 6x^2y^2-5xy-4

if we set t = xy, we have 6t2 - 5t - 4 = 0

then we need some method to factor the quadratic equation. What ways to do it you know?

we could use the quadratic formula t = [ -b ± √(b^2 - 4ac) ] / 2a
in our equation a = 6, b=-5, c=-4.

Could you do that? I really hate to take the joy of solving that from you, so I am going to be a bit sketchy on that, hoping that you'll ask more questions if you need more detailed explanation.
√(b^2 - 4ac)=√(25+4*4*6) =√(25+96) = √121 = 11

So after you put the numbers in the quadratic formula, you should arrive at t = 4/3 and t= -1/2 (please do check that!)

so 6(t-4/3)(t+1/2) and remember we put t=xy?

so 6(xy-4/3)(xy+1/2) is for (a) part of the problem

3z-8xyz+4x^2y^2z

we could factor z out right away. So

z(3-8xy+4x^2y^2)

now we do this substitution t=xy again. It is a very useful tool, problems look so much easier after substitution!!
z(3-8t+4t^2)

now we use quadratic formula again
what are your a, b and c in this situation? Remember that a is a coefficient in front of the squared term  as in front of t^2 and don't forget signs.

so I have
4z(xy-1/2)(xy-3/2)
what do you have?

36y^2-49(x+1)^2
as I said before this is the a^2-b^2 problem. We are aware that a^2-b^2=(a-b)(a+b),

so our a here would be 6y
our b 7(x+1)

so,
(6y-7(x+1))(6y+7(x+1))

or if you like it better
(6y-7x-7)(6y+7x+7)